NCEA Level 3 Chemistry (90700) 2012 – page 1 of 4
Assessment Schedule – 2012
Chemistry: Describe properties of aqueous systems (90700)
Evidence Statement
Q
ONE
(a)(i)
(ii)
(iii)
(b)(i)
(ii)
(iii)
(c)
Evidence
–
Achievement
+
HCl(aq) + H2O() → Cl (aq) + H3O (aq)
CH3NH2() + H2O  CH3NH3+(aq) + OH–(aq)
NH4Cl(s) → NH4+(aq) + Cl–(aq)
NH4+(aq) + H2O  NH3(aq) + H3O+(aq)
H3O+ = Cl– > OH–
CH3NH2 > OH– = CH3NH3+ > H3O+
Cl– > NH4+ > H3O+ = NH3 > OH–
Note: Do not
accept H+, states
are not required,
equation may be
in part (c).
Any TWO of:
•
(allow inclusion of H2O
and commas in place of >)
Either:
All FOUR equations correct.
Any TWO equations correct.
•
AND
•
Either:
Recognises reasons for pH
variation are due to production
of H3O+ / OH–
OR
Recognises conductivity is
related to the number of ions
in solution.
ONE answer to merit
level.
ALL species and order
correct for TWO solutions.
All species correct for TWO
solutions identified.
+
HCl is a strong acid, so fully reacts with water to produce a high [H3O ].
pH < 7.
NH4Cl is an acidic salt that completely dissociates into its ions producing
NH4+ and Cl–. NH4+ is a weak acid, so only partially reacts with water to
produce H3O+. pH < 7, but higher than HCl since NH4+ has a lower
[H3O+] than HCl.
CH3NH2 is a weak base, so only partially reacts with water to produce
OH– ions. pH > 7.
Electrical conductivity relates to the concentration of mobile charged
particles present. In the case of solutions, conductivity relates to the
number of ions present.
Both HCl and NH4Cl completely react with water to produce a high
concentration of ions, so their conductivity will be high (and equal).
Since CH3NH2 is a weak base, it partially reacts with water to produce
only a few ions in the solution, making it a poor electrical conductor.
•
OR
•
Achievement with
Excellence
Achievement with Merit
•
AND
Either:
Recognises reasons for
variations in pH and
conductivity AND makes a
valid comparison between
one pair.
OR
Difference in pH correctly
discussed for ALL 3
solutions.
OR
Difference in conductivity
correctly discussed for ALL
3 solutions.
•
Discussion addresses
variation in BOTH pH
(including whether acidic
or basic) and conductivity
using correct reasons for
ALL 3 solutions.
NCEA Level 3 Chemistry (90700) 2012 – page 2 of 4
Q
TWO
(a)(i)
Evidence
Fe(OH)2(s) → Fe2+(aq) + 2OH–(aq)
Achievement
(states not required,
Any TWO of:
allow )
•
(ii)
Ks = [Fe2+] [OH–]2
(b)
Let s be the solubility:
[Fe2+] = s
[OH–] = 2s
Ks = s × (2s)2
4.10 × 10–15 = 4s3
s = 1.01 × 10–5 mol L–1
Solubility of Fe(OH)2(s) = 1.01 × 10–5 mol L–1
(a) (i) and (ii) correct.
•
(c) (i)
•
When the pH is decreased, [H3O+] will increase. The H3O+ will react with
the OH– and therefore remove them from the equilibrium. This will cause
the reaction to replace some of the removed OH–. As a result more
Fe(OH)3 will dissolve, so decreasing the pH will increase the solubility of
Fe(OH)3.
Either:
Correct answer
(3 significant figures).
Method correct but error in
calculation.
OR
•
Fe(OH)3(s)  Fe3+(aq) + 3OH–(aq)
Ion Product (IP) = [Fe3+] [OH–]3
At pH 7, [OH–] = 1 × 10–7 mol L–1
IP = [1.05 × 10–4] [1 × 10–7]3 = 1.05 × 10–25
Since IP > Ks, Fe(OH)3 will form a precipitate
(ii)
Achievement with
Excellence
Achievement with Merit
•
Either:
Correct IP expression
OR
Compares IP and Ks to make a
valid conclusion.
Either:
Writes an equilibrium
expression AND identifies
direction it shifts in.
OR
States [OH–] decreases /
[H3O+] increases causing
Fe(OH)3 to be more soluble.
Method uses correct IP
expression but has one
calculation error AND
Compares IP and Ks to make
a valid conclusion
(3 significant figures).
•
AND
AND
•
Either:
States the change in [OH–],
its impact on the equilibrium
position and therefore more
Fe(OH)3 dissolves.
OR
Discussion of effect of
decreasing pH on Fe(OH)3
dissolving in terms of [H3O+]
/ [OH–] changing.
Answer correct with
supporting calculation
(3 significant figures).
•
Complete discussion of
effect of decreasing pH
on Fe(OH)3 solubility,
including role of H3O+
(reacting with OH–).
NCEA Level 3 Chemistry (90700) 2012 – page 3 of 4
Q
THREE
(a)
(b)
Evidence
NH3 + H2O  NH4+ + OH–
–9.24
–10
Ka = 10
= 5.75 × 10
Kb = [OH–]2 / [NH3] = Kw / Ka = 1.74 × 10–5
[OH–] = √(1.74 × 10–5 × 0.150) = 1.61 × 10–3 mol L–1
[H3O+] = Kw/[OH–] = 6.19 × 10–12 mol L–1
pH = 11.2
Achievement
Any TWO of:
• Either:
Ka correctly calculated
OR
Correct equilibrium equation
and Ka (or Kb) expression for
the equation written.
NH4+ + H2O  NH3 + H3O+
[NH4+] = [NH3] [H3O+] / Ka
[NH4+] = 0.150 × 10–8.60 / 10–9.24
[NH4+] = 0.655 mol L–1
n (NH4+) = 0.655 mol L–1 × 0.250 L = 0.164 mol
m (NH4Cl) = 0.164 mol × 53.5 g mol–1 = 8.76 g
Note: allow use of pH = pKa + log [NH3] / [NH4+]
•
Achievement with
Excellence
Achievement with Merit
Any TWO of:
•
Method correct with minor
arithmetic error.
•
Either:
pH correct (3 s.f.).
•
Ka expression or
pH = pKa + log[NH3] / [NH4+]
rearranged for [NH4+].
Either:
Correct method but error in
calculation / units missing /
unit incorrect
OR
[NH4+] calculated.
OR
Correct answer with units
(3 s.f.).
AND
(c)
A buffer is a solution that undergoes a minimal change of pH when small
amounts of acid or base are added.
Added acid will react with NH3 so that there is almost no change in
[H3O+]:
NH3 + H3O+ → NH4+ + H2O
Added base will react with NH4+ so that there is almost no change in
[OH–]:
NH4+ + OH– → NH3 + H2O
(These equations show that the ratio of NH3: NH4+ changes slightly, but
this does not significantly affect the pH.)
Since the pH of the buffer is lower than the pKa of NH4+, the [NH4+] will
be higher than the [NH3]. This means the buffer will be more effective
against added base.
•
Either:
Recognises acid will react
with added base and base will
react with added acid
OR
Describes the function of a
buffer.
•
Explains buffer action, writes
appropriate equations and
refers to the fact that there is
almost no change in [H3O+] /
[OH-].
•
Complete discussion of
buffer action and its
effectiveness at pH 8.60.
NCEA Level 3 Chemistry (90700) 2012 – page 4 of 4
Q
FOUR
(a)(i)
(ii)
(b)
(c)
Evidence
Achievement
n (NaOH) = 0.180 mol L–1 × 0.04 L = 0.0072 mol
1NaOH : 1HCOOH
c (HCOOH) = 0.0072 mol / 0.025 L = 0.288 mol L–1
Any TWO of:
HCOOH + H2O  HCOO– + H3O+
Ka = [HCOO–][H3O+] / [HCOOH]
[H3O+] = √(0.288 mol L–1 × 1.82 × 10–4) = 7.24 × 10–3 mol L–1
pH = 2.14
•
Shows [HCOOH] = 0.288
molL–1
•
Method to find pH correct with
minor arithmetic error.
Half way to the equivalence point (20 mL), the NaOH has reacted with half
the HCOOH, so [HCOOH] = [HCOO–].
Ka = [HCOO–] [H3O+] / [HCOOH]
Since [HCOOH] = [HCOO–], substitution into Ka gives:
Ka = [H3O+]
Taking –log of each side:
pKa = pH
So the pH half way to the equivalence point = pKa of HCOOH = 3.74
•
According to the titration curve, the pH at the equivalence point is
approximately 8.4. Indicators change colour at their pKa +/–1. So
bromocresol green changes colour over a pH range of 3.7-5.7 and alizarin
yellow will change colour over a pH range of 10.0 – 12.0. This means
bromocresol green would therefore change colour before the equivalence
point and alizarin yellow would change colour after the equivalence point.
Both of these indicators would therefore be unsuitable for this titration.
Cresol red will change colour over a pH range of 7.3 – 9.3. This includes
the pH at the equivalence point. Cresol red would therefore be a suitable
indicator to detect the endpoint since it changes colour at the equivalence
point.
•
Any TWO of:
•
Correct pH.
•
•
EITHER:
States [HCOOH] = [HCOO–]
EITHER:
Identifies correct or incorrect
indicator(s) with limited
reasoning.
OR
Identifies that the endpoint
needs to be close to the
equivalence point.
1 answer to merit level
States [HCOOH] =
[HCOO–] so pH = pKa when
half-way to the equivalence
point / in buffer zone.
OR
pH = pKa
Achievement with
Excellence
Achievement with Merit
AND
•
Identifies the correct
indicator and an incorrect
indicator, including
consequences of wrong
choice, but gaps in
discussion (i.e. have talked
in general terms about an
incorrect choice).
•
Discussion of
indicators links the
equivalence point pH
to the effective pKa
over which each
indicator will change
colour (range may be
for either the indicator
colour change
occurring over a range
OR for it’s pKa falling
within the range of the
vertical part of graph).
Judgement Statement
Achievement
3×A
or
2×M
Achievement with Merit
Achievement with Excellence
3×M
2×E+1×M